Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Help me to Understand CFL circuit

Status
Not open for further replies.

chul

Newbie level 3
Newbie level 3
Joined
May 6, 2011
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,307
I am struggling to understand CFL circuit. I have done it half way and need your support to complete !!!
Dear friends, can you tell me how the
1. oscillating circuit works - how the transistors or FET's switching done(control).
2. how the resonance frequency generates for resonance tank at the start.
3. what are the current paths when oscillating.

It's ok if you answer with a circuit as your preference or use this link.

Thanks a lot in advance.... 8O
 

The operation is basically the same as a low frequency magnetic ballast, just raised in frequency to make the ballast smaller.

Fluorescent bulb do work more efficient at between 25 kHz and 45 kHz drive frequency.

Rectify AC main to create HV DC, chop the DC to create 30 kHz squarewave, add a series powered iron core inductor as ballast. The starting is a small cap or diac between end filiments of bulb to create a transient voltage boost by 'twanging' inductor when power is first applied to ignite bulb.

Ballast inductance sets the run current of the bulb based on the bulb wattage, drive frequency, and HV square wave voltage.

Higher wattage bulb sometimes have an additional boost pre-switcher to provide a power factor correction on charging the HV DC filter cap.

The actual operation of a Fluorescent bulb is quite complicated. It is a negative resistance slope device. Gas in the bulb is ionized to create UV radiation that causes fluorescent powder coating on inside of bulb to emit visible light. There is a small amount of mercury that vaporizes creating the ionized gas. It takes a high voltage to strike the gas into ionization. The negative resistance slope means the voltage across the bulb goes down as the current through the bulb goes up.

They need to be fed from a current source. The current source approximation is a high voltage AC voltage that runs through an inductor ballast.
 
Last edited:
  • Like
Reactions: chul

    chul

    Points: 2
    Helpful Answer Positive Rating
My friend, i see you present the time-old circuit.

You see that as RCINFLA says, it is started off by diac.

but then the transistors are alternately switched due to saturation of drive transformer, so transistor is turned off by its drive winding saturating......then other transistro turns on, then its winding saturates, and it turns off, etc etc and so on.

I see there is no warm up circuit in your cct.

but yours is nice n cheap
 
  • Like
Reactions: chul

    chul

    Points: 2
    Helpful Answer Positive Rating
It is important to try and keep the duty cycle of the high frequency chopper as close to 50-50% duty as possible. Any DC component will imbalance the current flow in the bulb causing accelerated plating out of mercury and filiment material to one end of bulb (the black spots) reducing the bulb's life.

Many circuits have a cap in series with the bulb to remove DC component due to unbalanced duty cycle. This helps but there will still be a reduction in bulb life due to higher peak current in one direction.

The gas ionization characteristics has a negative temperature coefficient requiring a higher strike and run voltage at colder temps. Most CFL do not compensate for this which is why they work poorly at colder temps.

As bulb ages, more and more mercury is plated out of gas vapor reducing the ability of gas to ionize. This increases the strike and run voltages. Since most inexpensive CFL do not have any current sensing compensation this results in early failure of ability of bulb to strike into run state.
 
Last edited:
Thank you very much my dear friends., appreciate all of your information.
And ee2am:
but then the transistors are alternately switched due to saturation of drive transformer, so transistor is turned off by its drive winding saturating......then other transistro turns on, then its winding saturates, and it turns off, etc etc and so on.

This is what I wanted exactly , But in detail. I see the dot convention helps opposed biasing to Transistors, but I can't understand, how the driver transformers core saturation helps to switch the voltage build up direction. I studied B-H curve but yet i can't figure how it goes to it's negative cycle and create a voltage in opposite direction.
Kindly request your assistance :)
 

But in detail. I see the dot convention helps opposed biasing to Transistors, but I can't understand, how the driver transformers core saturation helps to switch the voltage build up direction. I studied B-H curve but yet i can't figure how it goes to it's negative cycle and create a voltage in opposite direction.
It basically simple. The core saturation causes the transformer voltage to drop to zero although the primary current is maintained. The transistor will turn off, causing the primary current to decrease. This induces a voltage of opposite sign, turning on the other transistor. Current feedback makes the transistor fully turn on.
 
  • Like
Reactions: chul

    chul

    Points: 2
    Helpful Answer Positive Rating
Ahh haa.. here it comes :) can you kindly provide me some literature on this,, I tried so hard - but couldn't find good reading to understand it..
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top